Non linear curve fitting by changing parameters

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Araz
Araz 2012 年 11 月 19 日
Hi
Im trying to fitt two non linear curves to each other the red curve to the blue . I have tried the basic fitting method but it doesn't give me a good match.
In the equation h=... of my code by changing some parameters which is 2.41, 0.389, 0.3, 0.406, 0.06, 0.406 i should get the red curve to be fitted on blue curve.
The equation h in my code is :
h=sqrt(1+(2.41./(1-(0.389e-6./lambda).^2))+(0.3./((1-(0.406e-6./lambda).^2)+((1i.*0.06).*(0.406e-6./lambda))))); I have changed much the parameters but the curves doesn't fit.
What can i do to fit this two curves to each other by changing only thus six parameters?
Best Regards
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Star Strider
Star Strider 2012 年 11 月 19 日
編集済み: Star Strider 2012 年 11 月 19 日
The value of h is complex. Do you want to fit the real(h) and imag(h) components separately, or do you want to fit abs(h)?
I also see only five parameters: [2.41; 0.389E-006; 0.3; 0.406E-006; 0.06].
This post ‘fitting of two curves and 6 unknown parameters’ also seems to quote the same equation. How to R and w fit? Should they be in your equation also?

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回答 (1 件)

Michael
Michael 2012 年 11 月 19 日
What is the blue curve? Usually a curve fitter algorithm requires a model (your h function) with specified parameter space (your six numbers), but also data to be fitted. Is the blue curve a set of (h,lambda) points?
In this format I'd suggest trying lsqcurvefit, you can specify your model h(lambda) in the argument, along with a starting point and your data.
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Araz
Araz 2012 年 11 月 19 日
blue curve is transmission vs lambda. The red curve in my code is also transmission vs lambda. but the curve differ from each other. is it possible to send the code to u email? u can have a look!

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